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$2Na+2H_2O\rightarrow 2Na^++2OH^-+H_2\uparrow$
$x x x 0,5x$
$2K+2H_2O\rightarrow 2K^++2OH^-+H_2\uparrow$
$y y y 0,5y$
$Ba+2H_2O\rightarrow Ba^{2+}+2OH^-+H_2\uparrow$
$z z 2z z$
$HCl\rightarrow H^++Cl^-$
$4a 4a 4a$
$H_2SO_4\rightarrow 2H^++SO_4^{2-}$
$a 2a a$
$H^++OH^-\rightarrow H_2O$
Theo bài ra ta có:
$\begin{cases}23x+39y+137z=8,94 \\ 0,5x+0,5y+z=\frac{2,688}{22,4} \\x+y+2z=4a+2a \end{cases}\Rightarrow a=0,04 mol$
$m_{hỗn hợp muối}=m_{Na^+}+
m_{K^+}+
m_{Ba^{2+}}+
m_{Cl^-}+
m_{SO_4^{2-}}$
$\Rightarrow
m_{hỗn hợp muối}= 8,94+35,5.(4.0,04)+96.0,04=8,94+5,68+3,84=18,46 gam$
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Trả lời 01-01-13 08:32 PM
viet130480
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